The bell-curve for Luck

[Bruno]

Originally Posted by Verjigorm

How do you people do this stuff?!

I make computers do it.

#!perl -w
use strict;
use Games::Rolldice 'roll';
use List::Util qw(sum max min);

my %results;
my $count = 1000000;

for (my $x =0;  $x<$count; $x++) {
   my $result = min(roll("3d6", '-sum' => 'true'),roll("3d6", '-sum' => 'true'),roll("3d6", '-sum' => 'true'),roll("3d6", '-sum' => 'true'));
   $results{$result}+= 1;
}

my %result_analysis;

foreach my $roll (sort keys %results){
   $result_analysis{$roll} = sprintf('%05f',($results{$roll}/$count));
   print "$roll:\t$result_analysis{$roll}\n";
}

[Bruno]

That really illustrates why I do it the "lazy" way - 19 lines of code (counting blank lines for readability), then I just run the stupid thing and stop thinking about it. Five minutes later, results pop out.

[bcd]

Originally Posted by Bruno

Good job I didn't do that then. :)

No, I was referring to Grouchy Chris' table with the two rerolls statistics in it; I just didn't make it very clear that this was what I did. :P

[Grouchy Chris]

Originally Posted by bcd

It seems a bit unfair to give probabilities for Luck on the premise "the first roll has already failed", side by side with probabilities for normal use without that premise. Properly therefore, the "Regular" column should be all zeros to be comparable. :P

I take your point, but I think it's useful as a benchmark for those who are used to thinking about their chances of beating, say, a 12 on one roll.

[mhd]

   my $result = min(roll("3d6", '-sum' => 'true'),roll("3d6", '-sum' => 'true'),roll("3d6", '-sum' => 'true'),roll("3d6", '-sum' => 'true'));

Why is roll invoked 4 times?

[Bruno]

Originally Posted by mhd

   my $result = min(roll("3d6", '-sum' => 'true'),roll("3d6", '-sum' => 'true'),roll("3d6", '-sum' => 'true'),roll("3d6", '-sum' => 'true'));

Why is roll invoked 4 times?

Because the last time I rand it, it was for Ts_ - who asked for a "best of 4" scenario :)

[mhd]

Originally Posted by Bruno

Because the last time I rand it, it was for Ts_ - who asked for a "best of 4" scenario :)

Missed that, sorry ;)

[Dragondog]

Getting an 18 in best of 4 happens once in about 2.18 billion rolls. And getting 17 is 15 times more likely.

Originally Posted by Bruno

Actually, I double checked - since I got an 18 at all in my output, it means that at least one 18 was rolled. I don't register the number in the hash table unless it comes up.

For example, I just tested the "best of 4" scenario and no 17s or 18s ever came up :)

	4x Luck	3x Luck	Regular
3:	1.8375%	1.3832%	0.5%
4:	5.3775%	4.0677%	1.389%
5:	10.0936%	7.8090%	2.740%
6:	14.9343%	12.0218%	4.616%
7:	18.5378%	15.8806%	6.945%
8:	19.2047%	18.2009%	9.736%
9:	14.7819%	16.2288%	11.564%
10:	8.9922%	11.9093%	12.473%
11:	4.2772%	7.2197%	12.509%
12:	1.5167%	3.5351%	11.649%
13:	0.3788%	1.3187%	9.736%
14:	0.0607%	0.3452%	6.920%
15:	0.0069%	0.0706%	4.626%
16:	0.0002%	0.0089%	2.795%
17:		0.0006%	1.391%
18:		0.0000%	0.458%

[Verjigorm]

Originally Posted by Bruno

I make computers do it.

Could you do that with say, 4 d6, drop the lowest 2, as well as 5d6, drop lowest 2? what about roll 5 and drop the lowest three, roll 6 and drop the lowest three?

[PseudoFenton]

Originally Posted by Verjigorm

Could you do that with say, 4 d6, drop the lowest 2, as well as 5d6, drop lowest 2? what about roll 5 and drop the lowest three, roll 6 and drop the lowest three?

Is that a question or a request? It's possible, if that's what your question is.

Also, AnyDice.com does some fun statistical read outs for various die rolling methods and (as the name suggests) even fictional sided die, so you can check that out too.